Measures of central tendency — mean (grouped - ungrouped), median (grouped), mode (grouped): Question bank

True or false — justify

The mean is always one of the actual data values.

False. The mean is a balance point, not a member of the data — e.g. mean of

3, 4

3.5

, which never occurs. It lives between values.

The median must be one of the actual data values.

False for even

n

(it averages the two middle values, e.g.

5.5

) and false for grouped data (interpolation lands anywhere inside the median class). It is a member only when

n

is odd and data is ungrouped.

Adding a huge outlier changes the median a lot.

False. The median only cares about position in the sorted list, so one extreme value barely shifts the middle; it is the mean that gets dragged toward the outlier.

If all frequencies are equal, the grouped mode formula gives the modal class centre.

False — with equal frequencies there is no single peak;

f_{1} - f_{0}

and

f_{1} - f_{2}

can be zero, making the formula undefined or meaningless. The distribution has no mode.

For a perfectly symmetric distribution, mean, median and mode coincide.

True. Symmetry means the balance point, the middle, and the peak all sit at the axis of symmetry, so all three land on the same value.

The empirical relation

Mode \approx 3 Median - 2 Mean

is an exact identity.

False. It is an approximation valid only for moderately skewed unimodal data; for strongly skewed or multi-modal data it can be far off.

Doubling every observation doubles the mean, median and mode.

True. All three are linear location measures, so scaling every value by a constant

c

scales each measure by

c

(they track the shifted/stretched picture).

Adding 5 to every observation adds 5 to the mean but leaves the median unchanged.

False. Adding a constant shifts everything rightward by 5, so mean, median and mode all increase by 5 together.

The class with the largest cumulative frequency is the modal class.

False. Cumulative frequency only ever grows, so its largest value is always the last class. The modal class uses the largest individual frequency

f_{i}

Spot the error

A student writes the median class as "the class containing the

\frac{n + 1}{2}

-th item". Fix it.

The

\frac{n + 1}{2}

rule is the ungrouped median position. For grouped data we use

\frac{n}{2}

(continuous interpolation), locating the first class whose cumulative frequency

\geq \frac{n}{2}

Classes are

20 - 29, 30 - 39

and a student takes

l = 30

for the second class. What's wrong?

These are inclusive limits with a gap between 29 and 30. You must convert to continuous boundaries first (

29.5 - 39.5

), so the true lower boundary is

l = 29.5

, not

30

In the mode formula a student uses

f_{1} - f_{0}

in the denominator too, writing

\frac{f _{1} - f _{0}}{f _{1} - f _{0}}

. Why is that wrong?

The denominator must be the total excess over both neighbours,

2 f_{1} - f_{0} - f_{2} = (f_{1} - f_{0}) + (f_{1} - f_{2})

. Using only one side ignores the pull from the class after the peak.

To find the median, a student takes the middle of the list

7, 2, 9, 4

\frac{9 + 4}{2}

. Spot the mistake.

The list wasn't sorted. Order first:

2, 4, 7, 9

, then the middle two are

4

and

7

, giving median

5.5

— not

6.5

A student computes grouped mean using class limits (20, 30) instead of midpoints. Why does this bias the answer?

The mean assumes every value in a class sits at its centre. Using a limit systematically shifts every class's assumed value to one edge, dragging the whole mean too low (or too high).

Someone claims the step-deviation method gives a different mean than the direct method. Correct them.

It gives the identical mean. Step-deviation is just algebra (

x_{i} = a + h u_{i}

) rearranged to make arithmetic easier — no information is lost or approximated.

Why questions

Why does the grouped median formula use

\frac{h}{f}

as the "width per item"?

We assume the

f

items in the median class are spread evenly across width

h

, so each item occupies

\frac{h}{f}

of that width; multiplying by how many items we still need past the boundary tells us how far to step in.

Why can a mean be non-representative of a skewed data set?

A few extreme values pull the balance point toward the long tail, so the "fair share" no longer describes a typical observation — this is why median or mode is often preferred for skewed data. See Skewness.

Why do we assume every value in a class equals its midpoint for the grouped mean?

We've lost the individual values, so the midpoint is the least-biased guess: on average, values above and below the centre cancel, keeping the estimate unbiased for roughly symmetric classes.

Why does the mode formula lean the peak left when the previous class is heavier (

f_{0} > f_{2}

A heavier class before means more "mass" just left of the peak, so the smooth peak of the underlying shape sits closer to the left edge of the modal class — the fraction

\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}

shrinks toward zero.

Why is the median found from an ogive at the height

\frac{n}{2}

The ogive plots cumulative frequency; reading across at half the total finds the value where exactly half the data lies below — precisely the median's definition. See Cumulative Frequency & Ogives.

Why does setting

\sum (x_{i} - m) = 0

define the balance point?

It says the total "pull" of values above

m

exactly cancels the total pull below, which is the physical condition for a see-saw to balance at

m

— solving it gives

m = \frac{\sum x _{i}}{n}

Why do all three measures agree for a symmetric distribution but split apart under skew?

Symmetry forces the peak, middle and balance point onto one axis; skew stretches one tail, moving the mean furthest (it feels the tail), the mode least (it only sees the crowd), with the median in between. See Skewness.

Edge cases

What is the mode of the data

2, 3, 4, 5

where every value appears once?

There is no mode — no value is more crowded than another. A data set can have zero, one, or several modes.

If a data set is

1, 1, 2, 2, 3, 3

, how many modes are there?

Three modes (

1, 2, 3

) — it is multimodal. The single-peak grouped mode formula is invalid here since there's no unique modal class.

The modal class is the first class in the table, so there is no "class before". What do you do?

Treat the missing

f_{0}

0

. The formula still works: the peak simply can't lean further left than the lower boundary

l

The median falls exactly on a class boundary (i.e.

c f

before equals

\frac{n}{2}

). What does the formula give?

The numerator

\frac{n}{2} - c f = 0

, so Median

= l

exactly — the middle person sits right at the lower boundary, which the formula respects cleanly.

For a single data point

{7}

, what are the mean, median and mode?

All equal

7

. With

n = 1

the balance point, middle, and only value coincide trivially — a useful sanity check for any formula.

What happens to the mode formula if

f_{1} = f_{0} = f_{2}

(flat top)?

The denominator

2 f_{1} - f_{0} - f_{2} = 0

, so the formula is undefined — there is no rise to locate a peak, matching the fact that a flat region has no single mode.

If two classes tie for the highest frequency, which is the modal class?

Neither uniquely — the distribution is bimodal and the single-peak formula shouldn't be applied. Report both peaks or use a finer grouping to break the tie.

Recall Quick self-test

The mean is dragged by outliers but the median isn't — why? ::: The mean sums values (an outlier contributes its full magnitude), while the median counts positions (an outlier is just "one more item at the end"). A distribution with no repeated values has what mode? ::: No mode — nothing is more frequent than anything else. Whose lower boundary do the median AND mode formulas need — limit or boundary? ::: The continuous boundary, never the printed inclusive limit.